Question

Prob 5 You have a multiple choice question with 4 choices. If you know the answer, you get the question right of course. If you do not know the answer, you pick a choice at random. Given you got an answer right, what is the probability you knew the answer? -solution to prob 5- Prob 5 Let K be the event that you know the answer, C the event that you get the answer correct. The event you guess is the event you do not know, i.e., Ke. Furthermore, P(CIK) 1 and P(CiK The probability of K given C requires us to invert the order the conditioning. So we use Bayes' rule P(KC)P(C) P(C) We expand the denominator using the rule of total probability (K and Ke form a partition of the sample space), P(CIK)P(K P(CK)P(K) P(K) The sample space with events K denotes the probability you knew the answer and R denotes the probability the answer was right. You guess if you do not know, ie, the event you guess is Ke We saw that the conditional probability P(KIR) depends on P(K). The quantity P(K) is usually called the prior probability (as in prior to the observation of whether the answer was right/wrong), and the quantity P(KIR) is called the posterior (after we make the observation) 1. If you guess all the time (whether or not you know the answer), do you still need P(K) to compute P(KR)? 2. Do you need P(K) to compute P(K|Rc)? If not, what is the numerical value of P(K|Rc)? Does it make sense? 3. Suppose there are two students, Focused (F) and Distracted (D). The student F has P(K) 3/4 and D has probability P(K) 1/3 (and their probabilities are independent). But F is so focused on the problem and D is so distracted by other things that they both forget to write their names on their exams! Suppose you find one exam with the correct answer and one with the wrong one. What is the conditional probability that F got it wrong and D got it correct?

Answer 1

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